Problems On Trains - Aptitude MCQ Questions and Solutions with Explanations

$$\eqalign{ & {\text{Let the speed of first train be }} \cr & {{\text{S}}_1}{\text{ km/hr and speed of second train}} \cr & {\text{is }}{{\text{S}}_2}{\text{km/hr }} \cr & {\text{As we know,}} \cr & {\text{Time }} \cr & {\text{ = }}\frac{{{\text{total distance}}}}{{{\text{relative speed in same/opposite direction}}}} \cr & {\text{In the same direction}} \cr & \Rightarrow {\text{27 sec = }}\frac{{\left( {100 + 95} \right)}}{{\left( {{\text{ }}{{\text{S}}_1} - {\text{ }}{{\text{S}}_2}} \right) \times \frac{5}{{18}}}} \cr & \Rightarrow 27 = \frac{{195 \times 18}}{{\left( {{\text{ }}{{\text{S}}_1} - {\text{ }}{{\text{S}}_2}} \right) \times 5}} \cr & \Rightarrow {\text{ }}{{\text{S}}_1} - {\text{ }}{{\text{S}}_2} = 26.......................(i) \cr & {\text{In the opposite direction,}} \cr & \Rightarrow 9 = \frac{{\left( {100 + 95} \right)}}{{\left( {{\text{ }}{{\text{S}}_1}{\text{ + }}{{\text{S}}_2}} \right) \times \frac{5}{{18}}}} \cr & \Rightarrow 9 = \frac{{195 \times 18}}{{\left( {{\text{ }}{{\text{S}}_1}{\text{ + }}{{\text{S}}_2}} \right) \times 5}} \cr & \Rightarrow {\text{ }}{{\text{S}}_1}{\text{ + }}{{\text{S}}_2} = 39 \times 2 \cr & \Rightarrow {\text{ }}{{\text{S}}_1}{\text{ + }}{{\text{S}}_2} = 78 \cr & {\text{From equation (i) and (ii)}} \cr & \Rightarrow {\text{ }}{{\text{S}}_1} - {\text{ }}{{\text{S}}_2} = 26 \cr & \Rightarrow {\text{ }}{{\text{S}}_1}{\text{ + }}{{\text{S}}_2} = 78 \cr & \Rightarrow {\text{ }}{{\text{S}}_1} = \frac{{26 + 78}}{2} \cr & \Rightarrow {\text{ }}{{\text{S}}_1} = \frac{{104}}{2} \cr & \Rightarrow {\text{ }}{{\text{S}}_1}{\text{ = 52 km/hr and }} \cr & \,\,\,\,\,\,\,\,\,\,{{\text{S}}_2}{\text{ = 26 km/hr}} \cr} $$

$$\eqalign{ & {\text{Let length of train }} \cr & {\text{ = }}l\,{\text{metre}} \cr & \Rightarrow {\text{Time }} \cr & {\text{ = }}\frac{{{\text{total distance}}}}{{{\text{relative speed in opposite direction}}}} \cr & \Rightarrow 4\sec \, = \,\frac{{l + 0}}{{\left( {84 + 6} \right) \times \frac{5}{{18}}{\text{m/s}}}} \cr & \Rightarrow 4\, = \frac{l}{{90 \times \frac{5}{{18}}}} \cr & \Rightarrow \,l\, = \,100\,{\text{m}} \cr & \therefore {\text{ length of the train = 100 m}} \cr} $$

$$\eqalign{ & {\text{Let the length of train }}x{\text{ m }} \cr & {\text{Speed of train }} \cr & {\text{ = }}\frac{{\left( {{\text{Length of train + length of bridge }}} \right)}}{{{\text{Time taken in crossing}}}}{\text{ }} \cr & {\text{According to information we get}} \cr & \Rightarrow \frac{{x + 500}}{{100}} = \frac{{x + 250}}{{60}} \cr & \Rightarrow 60\left( {x + 500} \right) = 100\left( {x + 250} \right) \cr & \Rightarrow 3\left( {x + 500} \right) = 5\left( {x + 250} \right) \cr & \Rightarrow 5x + 1250 = 3x + 1500 \cr & \Rightarrow 5x - 3x = 1500 - 1250 \cr & \Rightarrow 2x = 250 \cr & \Rightarrow x = \frac{{250}}{2} = 125\,{\text{m}} \cr} $$

$$\eqalign{ & {\text{Let the length of train be x m}} \cr & {\text{When a train crosses a light }} \cr & {\text{post in 3 second the distance covered}} \cr & {\text{ = length of train }} \cr & \Rightarrow {\text{speed of train = }}\frac{x}{3} \cr & {\text{Distance covered in crossing a}} \cr & {\text{900 meter platfrom in 30 seconds}} \cr & {\text{ = Length of platfrom + length of train}} \cr & {\text{Speed of train = }}\frac{{x + 900}}{30} \cr & \Rightarrow \frac{x}{3} = \frac{{x + 900}}{{30}}\left[ {\because {\text{Speed = }}\frac{{{\text{Distance}}}}{{{\text{Time}}}}} \right] \cr & \Rightarrow \frac{x}{1} = \frac{{x + 900}}{{10}} \cr & \Rightarrow 10x = x + 900 \cr & \Rightarrow 10x - x = 900 \cr & \Rightarrow 9x = 900 \cr & \Rightarrow x = \frac{{900}}{9} = 100{\text{m}} \cr & {\text{When the length of the platform be 800m,}} \cr & {\text{then time T be taken by train to cross 800m}} \cr & {\text{long platfrom}} \cr & \frac{x}{3} = \frac{{x + 800}}{T} \cr & \Rightarrow Tx = 3x + 2400 \cr & \Rightarrow 100T = 300 + 2400 \cr & \Rightarrow 100T = 2700 \cr & \Rightarrow T = \frac{{2700}}{{100}} = 27{\text{ seconds}} \cr} $$

$$\eqalign{ & {\text{Given , }} \cr & {\text{Train cover 3584 kms in 2 days 8 hours}} \cr & \left( {2\,{\text{days 8 hours = }}\frac{7}{3}{\text{ days}}} \right) \cr & {\text{Average speed = }} {\frac{{3584}}{{ {\frac{{7}}{{3}}}}}} \cr & {\text{ = 1536 km/day = }}\frac{{1536}}{{24}}{\text{ = 64 km/h}} \cr & {\text{Distance covered in two days}} \cr & {\text{ = 1440 + 1608 = 3048 km}} \cr & {\text{Remaining distance for third day}} \cr & {\text{ = 3584 }} - {\text{3048 = 536 km}} \cr & {\text{Third day 536 km is covered in }} \cr & {\text{8 hour with speed of}} \cr & {\text{ = }}\frac{{536}}{8} = 67{\text{ km/h }} \cr & {\text{( 3rd day total 536 km distance}} \cr & {\text{ covered by 67 km/hr in 8 hr)}} \cr & \therefore {\text{Difference of average speedm}} \cr & {\text{ = 67}} - {\text{64 = 3 km/hr}} \cr} $$

The speed of train A is 50km/hr and A starts its journey at 7 AM and reaches C at 10 AM. Total Travel time = 3hr
∴ Distance cover by A in 3hr = 50 × 3 = 150KM
Similarly, the speed of train B is 60km/hr and B starts its journey at 8 AM and reaches C at 10 AM. Total Travel time = 2hr
∴ Distance cover by B in 2hr = 60 × 2 = 120KM
The ratio of the distance between AC : BC
= 150 : 120
= 5 : 4

$$\eqalign{ & {\text{Let the length of train be x m}} \cr & {\text{When a train crosses a light }} \cr & {\text{post in 9 second the distance covered}} \cr & {\text{ = length of train }} \cr & \Rightarrow {\text{speed of train = }}\frac{x}{9} \cr & {\text{Distance covered in crossing a}} \cr & {\text{700 meter platfrom in 30 seconds}} \cr & {\text{ = Length of platfrom + length of train}} \cr & {\text{Speed of train = }}\frac{{x + 700}}{30} \cr & \Rightarrow \frac{x}{9} = \frac{{x + 700}}{{30}}\left[ {\because {\text{Speed = }}\frac{{{\text{Distance}}}}{{{\text{Time}}}}} \right] \cr & \Rightarrow \frac{x}{3} = \frac{{x + 700}}{{10}} \cr & \Rightarrow 10x = 3x + 2100 \cr & \Rightarrow 10x - 3x = 2100 \cr & \Rightarrow 7x = 2100 \cr & \Rightarrow x = \frac{{2100}}{7} = 300{\text{m}} \cr & {\text{When the length of the platform be 800m,}} \cr & {\text{then time T be taken by train to cross 800m}} \cr & {\text{long platform}} \cr & \frac{x}{9} = \frac{{x + 800}}{T} \cr & \Rightarrow Tx = 9x + 7200 \cr & \Rightarrow 300T = 2700 + 7200 \cr & \Rightarrow 300T = 9900 \cr & \Rightarrow T = \frac{{9900}}{{300}} = 33{\text{ seconds}} \cr} $$

$$\eqalign{ & {\text{Speed of train A}} \cr & {\text{ = 63 kmph}} \cr & {\text{ = }}\left( {\frac{{63 \times 5}}{{18}}} \right){\text{m/sec}} \cr & {\text{ = 17}}{\text{.5 m/sec}} \cr & {\text{Speed of train B}} \cr & {\text{ = 54 kmph}} \cr & {\text{ = }}\left( {\frac{{54 \times 5}}{{18}}} \right){\text{m/sec = 15 m/sec}} \cr & {\text{If the length of train A be }}x{\text{ metre,}} \cr & {\text{then}} \cr & {\text{Speed of train A}} \cr & {\text{ = }}\frac{{{\text{Length of train + length of platform}}}}{{{\text{Time taken in crossing}}}}{\text{ }} \cr & \Rightarrow 17.5 = \frac{{x + 199.5}}{{21}} \cr & \Rightarrow 17.5 \times 21 = x + 199.5 \cr & \Rightarrow 367.5 = x + 199.5 \cr & \Rightarrow x = 367.5 - 199.5 \cr & \Rightarrow 168\,{\text{metres}} \cr & {\text{Relative speed}} \cr & {\text{ = ( Speed train A + Speed train B)}} \cr & {\text{ = (17}}{\text{.5 + 15) m/sec}} \cr & {\text{ = 32}}{\text{.5 m/sec}} \cr & {\text{Required time}} \cr & {\text{ = }}\frac{{{\text{ Length of train A + Length of train B}}}}{{{\text{Relative speed }}}} \cr & = \left( {\frac{{168 + 157}}{{32.5}}} \right){\text{seconds}} \cr & = 10\,{\text{seconds}} \cr} $$

$$\eqalign{ & {\text{Average speed of train}} \cr & {\text{ = 40 km/hr}} \cr & {\text{Reach at its destination at on time }} \cr & {\text{New average speed of train}} \cr & {\text{ = 35 km/h}} \cr & {\text{Time = 15 minutes}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{ = }}\frac{{15}}{{60}}{\text{hours }} \cr & {\text{Then distance travelled}} \cr & {\text{ = }}\frac{{40 \times 35}}{{40 - 35}}{\text{ }} \times \frac{{15}}{{60}} \cr & {\text{ = }}\frac{{40 \times 35}}{5}{\text{ }} \times \frac{{15}}{{60}} \cr & {\text{ = 70}}\,{\text{km}} \cr} $$

$$\eqalign{ & {\text{Distance = Speed }} \times {\text{Time}} \cr & {\text{Distance covered by train with the}} \cr & {\text{speed of 30 kmph in 12 minutes is }} \cr & {\text{ = 30}} \times \frac{{12}}{{60}} = 6{\text{km}} \cr & {\text{Distance covered by the same train}} \cr & {\text{with the speed of 45 kmph in 8 minutes is }} \cr & {\text{ = 45}} \times \frac{8}{{60}} = 6{\text{km}} \cr & {\text{Average speed}} \cr & {\text{ = }}\frac{{{\text{total distance}}}}{{{\text{total time}}}}. \cr & \Rightarrow \frac{{\left( {6 + 6} \right){\text{km}}}}{{\left( {12 + 8} \right)\min }} = \frac{{12}}{{20}} \times 60 \cr & {\text{ = 36 kmph}} \cr} $$